CBSE Class 12 Mathematics Vector Algebra Assignment Set 05

Selected NCERT Questions

Question. Find the scalar components of the vector \( \vec{AB} \) with initial point \( A(2, 1) \) and terminal point \( B(-5, 7) \). 
Answer: Let \( \vec{AB} = (-5 - 2)\hat{i} + (7 - 1)\hat{j} = -7\hat{i} + 6\hat{j} \)
Hence, scalar components are \( -7, 6 \).
[Note: If \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \) then \( x, y, z \) are called scalar component and \( x\hat{i}, y\hat{j}, z\hat{k} \) are called vector component.]

Question. Find the unit vector in the direction of the vector \( \vec{a} = \hat{i} + \hat{j} + 2\hat{k} \).
Answer: Unit vector in the direction of vector \( \vec{a} \) is given by \( \hat{a} = \frac{\vec{a}}{|\vec{a}|} \)
Now, \( \quad |\vec{a}| = \sqrt{(1)^2 + (1)^2 + (2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \)
So, unit vector \( \quad \hat{a} = \frac{\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{6}} = \frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k} \)

Question. Find a vector in the direction of vector \( 5\hat{i} - \hat{j} + 2\hat{k} \) which has a magnitude of 8 units.
Answer: \( \quad \quad |\vec{a}| = \sqrt{(5)^2 + (-1)^2 + (2)^2} = \sqrt{25 + 1 + 4} = \sqrt{30} \)
The unit vector in the direction of vector \( \vec{a} \) is
\( \quad \quad \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{30}}(5\hat{i} - \hat{j} + 2\hat{k}) \)
Now, vector in the direction of \( \vec{a} \) having magnitude 8 units is
\( \quad \quad 8\hat{a} = \frac{8}{\sqrt{30}}(5\hat{i} - \hat{j} + 2\hat{k}) = \frac{40}{\sqrt{30}}\hat{i} - \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k} \)

Question. Find the position vector of a point \( R \) which divides the line joining two points \( P \) and \( Q \) whose position vectors are \( \hat{i} + 2\hat{j} - \hat{k} \) and \( -\hat{i} + \hat{j} + \hat{k} \) respectively, in the ratio 2 : 1
(i) internally
(ii) externally
Answer:  (i) Let \( R \) be the point which divides the line joining the point \( P \) and \( Q \) internally in the ratio 2 : 1.
The position vector of point \( R = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2\hat{j} - \hat{k})}{2 + 1} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k} \).
(ii) Let \( R \) be the point which divides the line joining the points \( P \) and \( Q \) externally in the ratio 2 : 1
\( = \frac{2(-\hat{i} + \hat{j} + \hat{k}) - 1(\hat{i} + 2\hat{j} - \hat{k})}{2 - 1} = \frac{-2\hat{i} + 2\hat{j} + 2\hat{k} - \hat{i} - 2\hat{j} + \hat{k}}{1} = -3\hat{i} + 3\hat{k} \)

Question. Find the projection of the vector \( \hat{i} - \hat{j} \) on the vector \( \hat{i} + \hat{j} \). 
Answer: Let \( \vec{a} = \hat{i} - \hat{j} \) and \( \vec{b} = \hat{i} + \hat{j} \)
Projection of \( \vec{a} \) on \( \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{(\hat{i} - \hat{j}) \cdot (\hat{i} + \hat{j})}{\sqrt{1 + 1}} = \frac{1 - 1}{\sqrt{2}} = 0 \)

Question. Find \( |\vec{x}| \), if for a unit vector \( \vec{a}, (\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12 \).
Answer: Sol. Here \( \quad |\vec{a}| = 1 \) and \( (\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12 \)
Now, \( \quad (\vec{x} - \vec{a})(\vec{x} + \vec{a}) = 12 \)
\( \implies \) \( \vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} - \vec{a} \cdot \vec{a} = 12 \)
\( \implies \) \( |\vec{x}|^2 + \vec{x} \cdot \vec{a} - \vec{x} \cdot \vec{a} - |\vec{a}|^2 = 12 \)
\( \implies \) \( |\vec{x}|^2 - |\vec{a}|^2 = 12 \)
\( \implies \) \( |\vec{x}|^2 - (1)^2 = 12 \)
\( \implies \) \( |\vec{x}|^2 = 12 + 1 \)
\( \implies \) \( |\vec{x}| = \sqrt{13} \)

Question. Using vectors, prove that the points \( (2, -1, 3), (3, -5, 1) \) and \( (-1, 11, 9) \) are collinear. 
Answer: Let \( A (2, -1, 3), B (3, -5, 1) \) and \( C (-1, 11, 9) \) are three points.
To show that \( A, B, C \) are collinear.
\( \therefore \quad \vec{AB} = (3 - 2)\hat{i} + (-5 + 1)\hat{j} + (1 - 3)\hat{k} = \hat{i} - 4\hat{j} - 2\hat{k} = \sqrt{1^2 + (-4)^2 + (-2)^2} = \sqrt{21} \)
\( \quad \quad \vec{BC} = (-1 - 3)\hat{i} + (11 + 5)\hat{j} + (9 - 1)\hat{k} = -4\hat{i} + 16\hat{j} + 8\hat{k} = \sqrt{(-4)^2 + (16)^2 + (8)^2} = 4\sqrt{21} \)
\( \quad \quad \vec{AC} = (-1 - 2)\hat{i} + (11 + 1)\hat{j} + (9 - 3)\hat{k} = -3\hat{i} + 12\hat{j} + 6\hat{k} = \sqrt{(-3)^2 + (12)^2 + (6)^2} = 3\sqrt{21} \)
\( \because \quad |\vec{AC}| + |\vec{AB}| = |\vec{BC}| \)
\( \implies \) \( A, B, C \) are collinear.

Question. Find \( \lambda \) and \( \mu \) if \( (2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda \hat{j} + \mu \hat{k}) = \vec{0} \).
Answer: Here \( \quad (2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda \hat{j} + \mu \hat{k}) = \vec{0} \)
\( \therefore \quad \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{vmatrix} = \vec{0} \)
\( \implies \) \( (6\mu - 27\lambda)\hat{i} - (2\mu - 27)\hat{j} + (2\lambda - 6)\hat{k} = 0 \)
\( \implies \) \( 6\mu - 27\lambda = 0, \quad 2\mu - 27 = 0 \text{ and } 2\lambda - 6 = 0 \)
\( \implies \) \( \mu = \frac{27}{2} \text{ and } \lambda = 3 \)

Question. Find a unit vector perpendicular to each of the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \), where \( \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - 2\hat{k} \). 
Answer: Given, \( \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - 2\hat{k} \)
\( \therefore \quad \vec{a} + \vec{b} = 4\hat{i} + 4\hat{j} \) and \( \vec{a} - \vec{b} = 2\hat{i} + 4\hat{k} \)
Now, vector perpendicular to \( (\vec{a} + \vec{b}) \) and \( (\vec{a} - \vec{b}) \) is
\( (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} = (16 - 0)\hat{i} - (16 - 0)\hat{j} + (0 - 8)\hat{k} = 16\hat{i} - 16\hat{j} - 8\hat{k} \)
\( \therefore \quad \) Unit vector perpendicular to \( (\vec{a} + \vec{b}) \) and \( (\vec{a} - \vec{b}) \) is given by
\( \pm \frac{(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})}{|(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|} = \pm \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{\sqrt{16^2 + (-16)^2 + (-8)^2}} = \pm \frac{8(2\hat{i} - 2\hat{j} - \hat{k})}{8\sqrt{2^2 + 2^2 + 1^2}} \)
\( = \pm \frac{2\hat{i} - 2\hat{j} - \hat{k}}{\sqrt{9}} = \pm \left( \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \right) = \pm \frac{2}{3}\hat{i} \mp \frac{2}{3}\hat{j} \mp \frac{1}{3}\hat{k} \)

Question. Let \( \vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}, \vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k} \) and \( \vec{c} = 2\hat{i} - \hat{j} + 4\hat{k} \). Find a vector \( \vec{d} \) which is perpendicular to both \( \vec{a} \) and \( \vec{b} \) and \( \vec{c} \cdot \vec{d} = 15 \).
Answer: The vector \( \vec{d} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \), so we have \( \vec{d} = \lambda (\vec{a} \times \vec{b}) \).
Now \( \quad \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} = 32\hat{i} - \hat{j} - 14\hat{k} \)
So, \( \quad \vec{d} = 32\lambda \hat{i} - \lambda \hat{j} - 14\lambda \hat{k} \)
We have \( \vec{c} \cdot \vec{d} = 15 \)
\( \implies \) \( (2\hat{i} - \hat{j} + 4\hat{k}) \cdot (32\lambda \hat{i} - \lambda \hat{j} - 14\lambda \hat{k}) = 15 \)
\( \implies \) \( 64\lambda + \lambda - 56\lambda = 15 \)
\( \implies \) \( 9\lambda = 15 \)
\( \implies \) \( \lambda = \frac{15}{9} = \frac{5}{3} \)
\( \therefore \quad \) Required vector \( \vec{d} = \frac{5}{3}(32\hat{i} - \hat{j} - 14\hat{k}) \)
Thus, \( \quad \vec{d} = 32 \times \frac{5}{3} \hat{i} - \frac{5}{3} \hat{j} - 14 \times \frac{5}{3} \hat{k} = \frac{160}{3} \hat{i} - \frac{5}{3} \hat{j} - \frac{70}{3} \hat{k} \).

Question. The scalar product of the vector \( \hat{i} + \hat{j} + \hat{k} \) with the unit vector along the sum of vectors \( 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \lambda \hat{i} + 2\hat{j} + 3\hat{k} \) is equal to one. Find the value of \( \lambda \).
Answer: Let sum of vectors \( 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \lambda \hat{i} + 2\hat{j} + 3\hat{k} \) equal to \( \vec{a} \) then \( \vec{a} = (2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k} \)
The unit vector in the direction of \( \vec{a} = \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 36 + 4}} \)
Hence, \( (\hat{i} + \hat{j} + \hat{k}) \cdot \hat{a} = (\hat{i} + \hat{j} + \hat{k}) \cdot \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 40}} = 1 \)
\( \implies \) \( (2 + \lambda) + 6 - 2 = \sqrt{(2 + \lambda)^2 + 40} \)
\( \implies \) \( (\lambda + 6)^2 = (2 + \lambda)^2 + 40 \)
\( \implies \) \( \lambda^2 + 36 + 12\lambda = 4 + \lambda^2 + 4\lambda + 40 \)
\( \implies \) \( 8\lambda = 8 \)
\( \implies \) \( \lambda = 1 \).

Question. If with reference to the right handed system of mutually perpendicular unit vectors \( \hat{i}, \hat{j} \) and \( \hat{k}, \vec{\alpha} = 3\hat{i} - \hat{j}, \vec{\beta} = 2\hat{i} + \hat{j} - 3\hat{k} \), then express \( \vec{\beta} \) in the form \( \vec{\beta} = \vec{\beta}_1 + \vec{\beta}_2 \), where \( \vec{\beta}_1 \) is parallel to \( \vec{\alpha} \) and \( \vec{\beta}_2 \) is perpendicular to \( \vec{\alpha} \).
Answer: Let \( \vec{\beta}_1 = \lambda \vec{\alpha}, \lambda \) is a scalar, i.e., \( \vec{\beta}_1 = 3\lambda \hat{i} - \lambda \hat{j} \)
and \( \vec{\beta}_2 = \vec{\beta} - \vec{\beta}_1 = (2 - 3\lambda)\hat{i} + (1 + \lambda)\hat{j} - 3\hat{k} \)
Now, since \( \vec{\beta}_2 \) is to be perpendicular to \( \vec{\alpha} \), we should have \( \vec{\alpha} \cdot \vec{\beta}_2 = 0 \) i.e.,
\( \quad \quad 3(2 - 3\lambda) - (1 + \lambda) = 0 \)
\( \implies \) \( \lambda = \frac{1}{2} \)
Therefore, \( \quad \vec{\beta}_1 = \frac{3}{2}\hat{i} - \frac{1}{2}\hat{j} \) and \( \vec{\beta}_2 = \frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k} \)

Very Short Answer Questions

Question. Find a vector in the direction of vector \( \vec{a} = \hat{i} - 2\hat{j} \) that has magnitude 7 units. 
Answer: The unit vector in the direction of the given vector \( \vec{a} \) is
\( \hat{a} = \frac{1}{|\vec{a}|} \vec{a} = \frac{1}{\sqrt{5}} (\hat{i} - 2\hat{j}) = \frac{1}{\sqrt{5}} \hat{i} - \frac{2}{\sqrt{5}} \hat{j} \)
Therefore, the vector having magnitude equal to 7 and in the direction of \( \vec{a} \) is
\( 7\hat{a} = 7 \left( \frac{1}{\sqrt{5}} \hat{i} - \frac{2}{\sqrt{5}} \hat{j} \right) = \frac{7}{\sqrt{5}} \hat{i} - \frac{14}{\sqrt{5}} \hat{j} \)

Question. Write the number of vectors of unit length perpendicular to both the vectors \( \vec{a} = 2\hat{i} + \hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{j} + \hat{k} \). 
Answer: Number of vectors of unit length perpendicular to both vectors = 2

Question. If \( |\vec{a}| = \sqrt{3}, |\vec{b}| = 2 \) and angle between \( \vec{a} \) and \( \vec{b} \) is \( 60^\circ \), then find \( \vec{a} \cdot \vec{b} \). 
Answer: Given, \( |\vec{a}| = \sqrt{3}, |\vec{b}| = 2 \) and \( \theta = 60^\circ \)
\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = \sqrt{3} \cdot 2 \cdot \cos 60^\circ = \sqrt{3} \quad \left[ \because \cos 60^\circ = \frac{1}{2} \right] \)

Question. Write the value of \( p \) for which \( \vec{a} = 3\hat{i} + 2\hat{j} + 9\hat{k} \) and \( \vec{b} = \hat{i} + p\hat{j} + 3\hat{k} \) are parallel vector.
Answer: Since \( \vec{a} \parallel \vec{b} \), therefore \( \vec{a} = \lambda \vec{b} \)
\( \implies \) \( 3\hat{i} + 2\hat{j} + 9\hat{k} = \lambda (\hat{i} + p\hat{j} + 3\hat{k}) \)
\( \implies \) \( \lambda = 3, 2 = \lambda p, 9 = 3\lambda \quad \text{or} \quad \lambda = 3, \; p = \frac{2}{3} \) \quad [By comparing the coefficients]

Question. Write a vector of magnitude 15 units in the direction of vector \( \hat{i} - 2\hat{j} + 2\hat{k} \).
Answer: Let \( \vec{a} = \hat{i} - 2\hat{j} + 2\hat{k} \)
Unit vector in the direction of \( \vec{a} \) is \( \hat{a} = \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{\sqrt{(1)^2 + (-2)^2 + (2)^2}} = \frac{1}{3}(\hat{i} - 2\hat{j} + 2\hat{k}) \)
Vector of magnitude 15 units in the direction of \( \vec{a} = 15 \hat{a} = 15 \frac{(\hat{i} - 2\hat{j} + 2\hat{k})}{3} \)
\( = \frac{15}{3}\hat{i} - \frac{30}{3}\hat{j} + \frac{30}{3}\hat{k} = 5\hat{i} - 10\hat{j} + 10\hat{k} \)

Question. What is the cosine of the angle, which the vector \( \sqrt{2}\hat{i} + \hat{j} + \hat{k} \) makes with y-axis?
Answer: We will consider \( \vec{a} = \sqrt{2}\hat{i} + \hat{j} + \hat{k} \)
Unit vector in the direction of \( \vec{a} \) is \( \hat{a} = \frac{\sqrt{2}\hat{i} + \hat{j} + \hat{k}}{\sqrt{(\sqrt{2})^2 + (1)^2 + (1)^2}} = \frac{\sqrt{2}\hat{i} + \hat{j} + \hat{k}}{\sqrt{4}} = \frac{\sqrt{2}\hat{i} + \hat{j} + \hat{k}}{2} \)
\( = \frac{\sqrt{2}}{2}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k} \)
The cosine of the angle which the vector \( \sqrt{2}\hat{i} + \hat{j} + \hat{k} \) makes with y-axis is \( \left(\frac{1}{2}\right) \).

Question. If \( |\vec{a}| = 4, |\vec{b}| = 3 \) and \( \vec{a} \cdot \vec{b} = 6\sqrt{3} \), then the value of \( |\vec{a} \times \vec{b}| \). 
Answer: We have,
\( \vec{a} \cdot \vec{b} = 6\sqrt{3} \)
\( \implies \) \( |\vec{a}| |\vec{b}| \cos \theta = 6\sqrt{3} \)
\( \implies \) \( 4 \times 3 \cos \theta = 6\sqrt{3} \)
\( \implies \) \( \cos \theta = \frac{6\sqrt{3}}{4 \times 3} = \frac{\sqrt{3}}{2} \)
\( \implies \) \( \theta = \frac{\pi}{6} \)
Now, \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta = 4 \times 3 \sin \frac{\pi}{6} = 4 \times 3 \times \frac{1}{2} = 6 \)

Question. Find the sum of the vectors \( \vec{a} = \hat{i} - 2\hat{j} + \hat{k}, \vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k} \) and \( \vec{c} = \hat{i} - 6\hat{j} - 7\hat{k} \). 
Answer: \( \vec{a} + \vec{b} + \vec{c} = (1 - 2 + 1)\hat{i} + (-2 + 4 - 6)\hat{j} + (1 + 5 - 7)\hat{k} = -4\hat{j} - \hat{k} \)

Question. Write the value of the area of the parallelogram determined by the vectors \( 2\hat{i} \) and \( 3\hat{j} \).
Answer: Required area of parallelogram \( = |2\hat{i} \times 3\hat{j}| = 6 |\hat{i} \times \hat{j}| = 6 |\hat{k}| = 6 \text{ sq units.} \)
[Note: Area of parallelogram whose sides are represented by \( \vec{a} \) and \( \vec{b} \) is \( |\vec{a} \times \vec{b}| \)]

Question. Write the value of \( (\hat{i} \times \hat{j}) \cdot \hat{k} + (\hat{j} \times \hat{k}) \cdot \hat{i} \).
Answer: \( (\hat{i} \times \hat{j}) \cdot \hat{k} + (\hat{j} \times \hat{k}) \cdot \hat{i} = \hat{k} \cdot \hat{k} + \hat{i} \cdot \hat{i} = 1 + 1 = 2 \)
[Note: \( \vec{a} \cdot \vec{b} = |\vec{a}| \cdot |\vec{b}| \cos \theta \). Also \( |\hat{i}| = |\hat{j}| = |\hat{k}| = 1 \)]

Question. For what value of 'a' the vectors \( 2\hat{i} - 3\hat{j} + 4\hat{k} \) and \( a\hat{i} + 6\hat{j} - 8\hat{k} \) are collinear? 
Answer: \( \because 2\hat{i} - 3\hat{j} + 4\hat{k} \text{ and } a\hat{i} + 6\hat{j} - 8\hat{k} \text{ are collinear} \)
\( \therefore \frac{2}{a} = \frac{-3}{6} = \frac{4}{-8} \)
\( \implies \) \( a = \frac{2 \times 6}{-3} \text{ or } a = \frac{2 \times (-8)}{4} \)
\( \implies \) \( a = -4 \)
[Note: If \( \vec{a} \) and \( \vec{b} \) are collinear vectors then the respective components of \( \vec{a} \) and \( \vec{b} \) are proportional.]

Question. Write the direction cosines of the vector \( -2\hat{i} + \hat{j} - 5\hat{k} \).
Answer: Direction cosines of vector \( -2\hat{i} + \hat{j} - 5\hat{k} \) are
\( \frac{-2}{\sqrt{(-2)^2 + 1^2 + (-5)^2}}, \frac{1}{\sqrt{(-2)^2 + 1^2 + (-5)^2}}, \frac{-5}{\sqrt{(-2)^2 + 1^2 + (-5)^2}} = \frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}} \)
Note: If \( l, m, n \) are direction cosine of \( a\hat{i} + b\hat{j} + c\hat{k} \) then
\( l = \frac{a}{\sqrt{a^2+b^2+c^2}}, \quad m = \frac{b}{\sqrt{a^2+b^2+c^2}}, \quad n = \frac{c}{\sqrt{a^2+b^2+c^2}} \)

Question. If vectors \( \vec{a} \) and \( \vec{b} \) are such that \( |\vec{a}| = \frac{1}{2}, |\vec{b}| = \frac{4}{\sqrt{3}} \) and \( |\vec{a} \times \vec{b}| = \frac{1}{\sqrt{3}} \), then find \( |\vec{a} \cdot \vec{b}| \).
Answer: We have, \( |\vec{a} \times \vec{b}| = \frac{1}{\sqrt{3}} \)
\( \implies \) \( |\vec{a}| |\vec{b}| \sin \theta |\hat{n}| = \frac{1}{\sqrt{3}} \)
\( \implies \) \( |\vec{a}| |\vec{b}| \sin \theta = \frac{1}{\sqrt{3}} \quad [\because \theta \text{ is angle between } \vec{a} \text{ and } \vec{b}] \)
\( \implies \) \( \frac{1}{2} \times \frac{4}{\sqrt{3}} \cdot \sin \theta = \frac{1}{\sqrt{3}} \)
\( \implies \) \( \sin \theta = \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2} \)
\( \implies \) \( \sin \theta = \frac{1}{2} \)
\( \implies \) \( \theta = 30^\circ \)
Now, \( |\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| \cos \theta \)
\( = \frac{1}{2} \times \frac{4}{\sqrt{3}} \cdot \cos 30^\circ = \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} = 1 \)

Question. Let \( \vec{a} \) and \( \vec{b} \) be two vectors such that \( |\vec{a}| = 3 \) and \( |\vec{b}| = \frac{\sqrt{2}}{3} \) and \( \vec{a} \times \vec{b} \) is a unit vector. What is the angle between \( \vec{a} \) and \( \vec{b} \)?
Answer: We have, \( |\vec{a}| = 3, |\vec{b}| = \frac{\sqrt{2}}{3} \)
Now, \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \)
\( \implies \) \( \sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|} = \frac{1}{3 \times \frac{\sqrt{2}}{3}} = \frac{1}{\sqrt{2}} \)
\( \implies \) \( \theta = 45^\circ \)

Question. Give an example of vectors \( \vec{a} \) and \( \vec{b} \) such that \( |\vec{a}| = |\vec{b}| \) but \( \vec{a} \neq \vec{b} \).
Answer: Let \( \vec{a} = x\hat{i} + y\hat{j}; \quad \vec{b} = y\hat{i} + x\hat{j} \)
\( |\vec{a}| = \sqrt{x^2+y^2}, |\vec{b}| = \sqrt{y^2+x^2} \text{ Hence, } \vec{a} \neq \vec{b} \text{ but } |\vec{a}| = |\vec{b}| \)

Question. Find the unit vector in the direction of sum of vectors \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{j} + \hat{k} \). 
Answer: Let \( \vec{c} \) denote the sum of \( \vec{a} \) and \( \vec{b} \)
We have, \( \vec{c} = \vec{a} + \vec{b} = 2\hat{i} - \hat{j} + \hat{k} + 2\hat{j} + \hat{k} = 2\hat{i} + \hat{j} + 2\hat{k} \)
\( \therefore \text{Unit vector in the direction of } \vec{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{2^2 + 1^2 + 2^2}} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{9}} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{3} \)

Short Answer Questions

Question. The x-coordinate of a point on the line joining the point P(2, 2, 1) and Q(5, 1, -2) is 4. Find its z-coordinate.
Answer: Let required point be \( R(4, y_1, z_1) \) which divides \( PQ \) in ratio \( k : 1 \).
By section formula
\( 4 = \frac{5k + 2}{k + 1} \)
\( \implies \) \( 4k + 4 = 5k + 2 \)
\( \implies \) \( k = 2 \)
\( \therefore z_1 = \frac{2 \times (-2) + 1 \times 1}{2 + 1} = \frac{-4 + 1}{3} = \frac{-3}{3} = -1 \)

Question. Find '\( \lambda \)' when the projection of \( \vec{a} = \lambda\hat{i} + \hat{j} + 4\hat{k} \) on \( \vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k} \) is 4 units.
Answer: We know that projection of \( \vec{a} \) on \( \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \)
\( \implies \) \( 4 = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \quad ... (i) \)
Now, \( \vec{a} \cdot \vec{b} = 2\lambda + 6 + 12 = 2\lambda + 18 \) also \( |\vec{b}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{4 + 36 + 9} = 7 \)
Putting in (i), we get
\( 4 = \frac{2\lambda + 18}{7} \)
\( \implies \) \( 2\lambda = 28 - 18 \)
\( \implies \) \( \lambda = \frac{10}{2} = 5 \)

Question. What are the direction cosines of a line, which makes equal angles with the co-ordinate axes?
Answer: Let \( \alpha \) be the angle made by line with coordinate axes.
\( \implies \) \( \text{Direction cosines of line are } \cos \alpha, \cos \alpha, \cos \alpha. \)
\( \implies \) \( \cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1 \)
\( \implies \) \( 3 \cos^2 \alpha = 1 \)
\( \implies \) \( \cos^2 \alpha = \frac{1}{3} \)
\( \implies \) \( \cos \alpha = \pm \frac{1}{\sqrt{3}} \)
Hence, the direction cosines of the line equally inclined to the coordinate axes are
\( \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}} \)
[Note: If \( l, m, n \) are direction cosines of line then \( l^2+m^2+n^2=1 \)]

Question. For what value of p, is \( (\hat{i} + \hat{j} + \hat{k})p \) a unit vector?
Answer: Let, \( \vec{a} = p(\hat{i} + \hat{j} + \hat{k}) \)
Magnitude of \( \vec{a} = |\vec{a}| = \sqrt{(p)^2 + (p)^2 + (p)^2} = \pm\sqrt{3} p \)
As \( \vec{a} \) is a unit vector, \( |\vec{a}| = 1 \)
\( \implies \) \( \pm\sqrt{3} p = 1 \)
\( \implies \) \( p = \pm\frac{1}{\sqrt{3}}. \)

Question. X and Y are two points with position vectors \( 3\vec{a} + \vec{b} \) and \( \vec{a} - 3\vec{b} \) respectively. Write the position vector of a point Z which divides the line segment XY in the ratio 2 : 1 externally. 
Answer: We have \( \vec{OX} = 3\vec{a} + \vec{b}, \vec{OY} = \vec{a} - 3\vec{b}, \vec{OZ} = ? \)
\( \vec{OZ} = \frac{2(\vec{a} - 3\vec{b}) - 1(3\vec{a} + \vec{b})}{2 - 1} = \frac{-\vec{a} - 7\vec{b}}{1} = -\vec{a} - 7\vec{b} \)
\( \therefore \vec{OZ} = -\vec{a} - 7\vec{b} \)

Question. Write the position vector of the mid-point of the vector joining the points P(2, 3, 4) and Q(4, 1, -2). 
Answer: Let \( \vec{a}, \vec{b} \) be position vector of points P(2, 3, 4) and Q(4, 1, -2) respectively.
\( \therefore \vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} \text{ and } \vec{b} = 4\hat{i} + \hat{j} - 2\hat{k} \)
\( \therefore \text{Position vector of mid point of P and Q} = \frac{\vec{a} + \vec{b}}{2} = \frac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2} = 3\hat{i} + 2\hat{j} + \hat{k} \)

Question. If \( |\vec{a}| = a \), then find the value of the following : \( |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 \) 
Answer: Let \( \vec{a} \) makes angle \( \alpha, \beta, \gamma \) with x, y and z axis.
\( \therefore |\vec{a} \times \hat{i}| = |\vec{a}| \cdot 1 \cdot \sin \alpha = a \sin \alpha \text{ similarly } |\vec{a} \times \hat{j}| = a \sin \beta \text{ and } |\vec{a} \times \hat{k}| = a \sin \gamma \)
\( \therefore |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = a^2 \sin^2 \alpha + a^2 \sin^2 \beta + a^2 \sin^2 \gamma = a^2 [\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma] \)
\( = a^2 [1 - \cos^2 \alpha + 1 - \cos^2 \beta + 1 - \cos^2 \gamma] \)
\( = a^2 [3 - (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma)] \)
\( = a^2 (3 - 1) \quad \left[ \because l^2 + m^2 + n^2 = 1 \implies \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \right] \)
\( = 2a^2 \)

Question. The vectors \( \vec{a} = 3\hat{i} + x\hat{j} \) and \( \vec{b} = 2\hat{i} + \hat{j} + y\hat{k} \) are mutually perpendicular. If \( |\vec{a}| = |\vec{b}| \), then find the value of y. 
Answer: \( \because \vec{a} \text{ and } \vec{b} \text{ are mutually perpendicular.} \)
\( \therefore \vec{a} \cdot \vec{b} = 0 \)
\( \implies \) \( (3\hat{i} + x\hat{j}) \cdot (2\hat{i} + \hat{j} + y\hat{k}) = 0 \)
\( \implies \) \( 6 + x + 0 \cdot y = 0 \)
\( \implies \) \( 6 + x = 0 \)
\( \implies \) \( x = -6 \)
Again, \( |\vec{a}| = |\vec{b}| \)
\( \implies \) \( \sqrt{3^2 + x^2} = \sqrt{2^2 + 1 + y^2} \)
\( \implies \) \( \sqrt{9 + 36} = \sqrt{5 + y^2} \quad [\because x = -6] \)
\( \implies \) \( \sqrt{45} = \sqrt{5 + y^2} \)
\( \implies \) \( y^2 = 45 - 5 \)
\( \implies \) \( y = \pm\sqrt{40} = \pm 2\sqrt{10} \)

Question. Find the value of \( \vec{a} \cdot \vec{b} \) if \( |\vec{a}| = 10, |\vec{b}| = 2 \) and \( |\vec{a} \times \vec{b}| = 16 \). 
Answer: \( \because |\vec{a} \times \vec{b}| = 16 \)
\( \implies \) \( |\vec{a}||\vec{b}| \sin \theta = 16 \)
\( \implies \) \( 10 \times 2 \sin \theta = 16 \)
\( \implies \) \( \sin \theta = \frac{16}{20} = \frac{4}{5} \)
\( \implies \) \( \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{16}{25}} = \pm \frac{3}{5} \)
\( \therefore \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta = \pm 10 \times 2 \times \frac{3}{5} = \pm 12 \)

Question. Using vectors, find the value of k, such that the points (k, -10, 3), (1, -1, 3) and (3, 5, 3) are collinear. 
Answer: Let the points are A(k, -10, 3), B(1, -1, 3) and C(3, 5, 3).
So, \( \vec{AB} = \vec{OB} - \vec{OA} = (\hat{i} - \hat{j} + 3\hat{k}) - (k\hat{i} - 10\hat{j} + 3\hat{k}) = (1 - k)\hat{i} + 9\hat{j} + 0\hat{k} \)
\( \therefore |\vec{AB}| = \sqrt{(1 - k)^2 + (9)^2 + 0} = \sqrt{(1 - k)^2 + 81} \)
Similarly, \( \vec{BC} = \vec{OC} - \vec{OB} = (3\hat{i} + 5\hat{j} + 3\hat{k}) - (\hat{i} - \hat{j} + 3\hat{k}) = 2\hat{i} + 6\hat{j} + 0\hat{k} \)
\( \therefore |\vec{BC}| = \sqrt{2^2 + 6^2 + 0} = 2\sqrt{10} \)
and \( \vec{AC} = \vec{OC} - \vec{OA} = (3\hat{i} + 5\hat{j} + 3\hat{k}) - (k\hat{i} - 10\hat{j} + 3\hat{k}) = (3 - k)\hat{i} + 15\hat{j} + 0\hat{k} \)
\( \therefore |\vec{AC}| = \sqrt{(3 - k)^2 + 225} \)
If A, B and C are collinear, then sum of modulus of any two vectors will be equal to the modulus of third vectors i.e., \( |\vec{AB}| + |\vec{BC}| = |\vec{AC}| \),
\( \implies \) \( \sqrt{(1 - k)^2 + 81} + 2\sqrt{10} = \sqrt{(3 - k)^2 + 225} \)
\( \implies \) \( \sqrt{(3 - k)^2 + 225} - \sqrt{(1 - k)^2 + 81} = 2\sqrt{10} \)
\( \implies \) \( \sqrt{9 + k^2 - 6k + 225} - \sqrt{1 + k^2 - 2k + 81} = 2\sqrt{10} \)
\( \implies \) \( \sqrt{k^2 - 6k + 234} - 2\sqrt{10} = \sqrt{k^2 - 2k + 82} \)
\( \implies \) \( k^2 - 6k + 234 + 40 - 2\sqrt{k^2 - 6k + 234} \cdot 2\sqrt{10} = k^2 - 2k + 82 \)
\( \implies \) \( k^2 - 6k + 234 + 40 - k^2 + 2k - 82 = 4\sqrt{10}\sqrt{k^2 + 234 - 6k} \)
\( \implies \) \( -4k + 192 = 4\sqrt{10}\sqrt{k^2 + 234 - 6k} \)
\( \implies \) \( -k + 48 = \sqrt{10}\sqrt{k^2 + 234 - 6k} \)
On squaring both sides, we get
\( 48 \times 48 + k^2 - 96k = 10(k^2 + 234 - 6k) \)
\( \implies \) \( k^2 - 96k - 10k^2 + 60k = -48 \times 48 + 2340 \)
\( \implies \) \( -9k^2 - 36k = -48 \times 48 + 2340 \)
\( \implies \) \( (k^2 + 4k) = 16 \times 16 - 260 \quad [\text{dividing by 9 in both sides}] \)
\( \implies \) \( k^2 + 4k = -4 \)
\( \implies \) \( k^2 + 4k + 4 = 0 \)
\( \implies \) \( (k + 2)^2 = 0 \)
\( \therefore k = -2 \)

Question. Find the vector of magnitude 6, which is perpendicular to both the vectors \( 2\hat{i} - \hat{j} + 2\hat{k} \) and \( 4\hat{i} - \hat{j} + 3\hat{k} \).
Answer: Let \( \vec{a} = 2\hat{i} - \hat{j} + 2\hat{k} \text{ and } \vec{b} = 4\hat{i} - \hat{j} + 3\hat{k} \)
So, any vector perpendicular to both the vectors \( \vec{a} \) and \( \vec{b} \) is given by
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 2 \\ 4 & -1 & 3 \end{vmatrix} = \hat{i}(-3+2) - \hat{j}(6-8) + \hat{k}(-2+4) = -\hat{i} + 2\hat{j} + 2\hat{k} = \vec{r} \quad [\text{say}] \)
A vector of magnitude 6 in the direction of \( \vec{r} \)
\( = \frac{\vec{r}}{|\vec{r}|} \cdot 6 = \frac{-\hat{i} + 2\hat{j} + 2\hat{k}}{\sqrt{1^2 + 2^2 + 2^2}} \cdot 6 = \frac{-6}{3}\hat{i} + \frac{12}{3}\hat{j} + \frac{12}{3}\hat{k} = -2\hat{i} + 4\hat{j} + 4\hat{k} \)

Question. Let \( \vec{a} = \hat{i} + 2\hat{j} - 3\hat{k} \) and \( \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \) be two vectors. Show that the vectors \( (\vec{a} + \vec{b}) \) and \( (\vec{a} - \vec{b}) \) are perpendicular to each other. 
Answer: We have \( \vec{a} = \hat{i} + 2\hat{j} - 3\hat{k} \text{ and } \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}. \)
Then, \( \vec{a} + \vec{b} = 4\hat{i} + \hat{j} - \hat{k} \text{ and } \vec{a} - \vec{b} = -2\hat{i} + 3\hat{j} - 5\hat{k} \)
\( \because (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = -8 + 3 + 5 = 0 \)
\( \implies \) \( (\vec{a} + \vec{b}) \perp (\vec{a} - \vec{b}) \)

Question. For any two vectors \( \vec{a} \) and \( \vec{b} \), proved that \( (\vec{a} \times \vec{b})^2 = \vec{a}^2 \vec{b}^2 - (\vec{a} \cdot \vec{b})^2 \)
Answer: \( \because (\vec{a} \times \vec{b}) = |\vec{a}||\vec{b}| \sin \theta \hat{n} \text{ and } \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta \)
\( \text{RHS} = \vec{a}^2 \vec{b}^2 - (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 - |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta \)
\( = |\vec{a}|^2 |\vec{b}|^2 (1 - \cos^2 \theta) = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta \)
\( = (\vec{a} \times \vec{b})^2 = \text{LHS} \quad \text{Hence proved.} \)